Number of Jobs A sociologist found that in a random sample...

Number of Jobs A sociologist found that in a random sample of 50 retired men, the average number of jobs they had during their lifetimes was $7.2 .$ The population standard deviation is $2.1 .$ a. Find the best point estimate of the population mean. b. Find the $95 \%$ confidence interval of the mean number of jobs. c. Find the $99 \%$ confidence interval of the mean number of jobs. d. Which is smaller? Explain why.

Number of Jobs A sociologist found that in a random
sample of 50 retired men, the average number of jobs
they had during their lifetimes was $7.2 .$ The population
standard deviation is $2.1 .$
a. Find the best point estimate of the population
mean.
b. Find the $95 \%$ confidence interval of the mean number of jobs.
c. Find the $99 \%$ confidence interval of the mean number of jobs.
d. Which is smaller? Explain why.

Key Concepts

Critical Value

Critical values are derived from the standard normal distribution when the population standard deviation is known, and they correspond to the desired level of confidence. Higher confidence levels result in larger critical values, thus leading to wider confidence intervals.

Margin of Error

The margin of error quantifies the amount of random sampling error in the estimate. It is determined by the critical value associated with the desired confidence level, the population standard deviation, and the sample size, and it defines half the width of the confidence interval.

Confidence Interval

A confidence interval provides a range of values within which the true population parameter is likely to lie, with a specified level of confidence. It takes into account both the sample estimate and the uncertainty associated with the estimation process.

Point Estimation

This concept refers to using a sample statistic to estimate a population parameter. In this case, the sample mean is used as the best point estimate of the population mean because it is the most natural unbiased estimator under standard conditions.

Transcript

00:01 Okay, so in problem 12, we are given the following information.

00:06 A random sample of 50 retired men have an average number of jobs they had during their lifetime of 7 .2.

00:17 And the population's standard deviation is 2 .1.

00:22 Part a asks us to find the best point estimate of the population mean.

00:29 So the best point estimate of population mean is the sample meet.

00:38 So in this case, the best point estimate of the population mean is 7 .2 because that is the sample mean.

00:50 Part b says to find the 95 % confidence interval of the mean number of jobs.

00:58 So to do this, we need to know this formula right here.

01:08 X bar plus or minus z of alpha over 2 times sigma over the square root of n.

01:16 So we have everything that we need for this formula except for that z score.

01:24 So in order to find our z score, we need to find what alpha is.

01:30 So alpha is just one minus our confidence level, which in this case is 95%.

01:39 So it's going to be 1 minus 0 .95, which gives us 0 .05.

01:47 And since we're doing a confidence interval and we want both sides of the interval, we have to divide alpha by 2 and we get 0 .02.

02:01 So we need to find the z score of 0 .0 .025.

02:16 So using your calculator for your z tables, you can find what the z value is of an alpha.

02:24 025.

02:27 And you should find 1 .96 is your z score.

02:33 So once we have our z score and we have all the values we need to find the 95 % confidence interval.

02:47 So to find our confidence interval, we have our x bar, which is 7 .2 plus or minus our z score of 1 .96 times sigma over the square root of it.

03:11 I'm going to calculate this side first.

03:15 So we get 7 .2 plus or minus 0 .5821.

03:27 And then i can find the upper limit and the lower limit of my confidence interval.

03:39 So the upper limit is 7 .2 plus 0 .582.

03:48 And the lower limit is going to be 7 .2 minus 0 .582...

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